Approximating SHM Homework: F_\theta=-mg\theta

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Homework Help Overview

The discussion revolves around the approximation of the restoring force of a pendulum, specifically comparing the exact force \( F_\theta = -mg\sin\theta \) to its linear approximation \( F_\theta = -mg\theta \). Participants are examining the period of the pendulum, expressed as an infinite series, and questioning the nature of this approximation.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the series expansion of the period and questioning whether it represents a Maclaurin series. There is also discussion about the implications of large amplitude motion and the complexity of the resulting elliptical integral.

Discussion Status

The discussion is active, with participants providing insights and references to external resources. There is an ongoing examination of the variables involved in the pendulum's motion, particularly regarding the definitions and ranges of angles used in the equations.

Contextual Notes

Some participants note potential confusion regarding the variables used in the equations, specifically the roles of \( \alpha \), \( \varphi \), and \( \vartheta \) in describing the pendulum's motion. There is a recognition of the need for clarity on these definitions as they relate to the problem at hand.

Karol
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Homework Statement


The restoring force of a pendulum is F_\theta=-mg\sin\theta
and is approximated to F_\theta=-mg\theta.
The period is T=2\pi\sqrt{\frac{L}{g}}, but can be expressed as the infinite series:
T=2\pi\sqrt{\frac{L}{g}}\left( 1+\frac{1^2}{2^2}\sin^2\frac{\theta}{2}+\frac{1^2}{2^2}\frac{3^2}{4^2}\sin^4\frac{\theta}{2}+...\right)
What is this approximation and of what? i don't think it's a Maclaurin series.

Homework Equations


Maclaurin series of sin(x):
\sin(x)\cong 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...
 
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Karol said:

Homework Statement


The restoring force of a pendulum is F_\theta=-mg\sin\theta
and is approximated to F_\theta=-mg\theta.
The period is T=2\pi\sqrt{\frac{L}{g}}, but can be expressed as the infinite series:
T=2\pi\sqrt{\frac{L}{g}}\left( 1+\frac{1^2}{2^2}\sin^2\frac{\theta}{2}+\frac{1^2}{2^2}\frac{3^2}{4^2}\sin^4\frac{\theta}{2}+...\right)
What is this approximation and of what? i don't think it's a Maclaurin series.

Homework Equations


Maclaurin series of sin(x):
\sin(x)\cong 1-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...
When solving for the period of a large amplitude pendulum you get a nasty elliptical integral. You can express that integral in terms of a series, the first term being the familiar period for small amplitudes.

Try this for more details: LARGE-ANGLE MOTION OF A SIMPLE PENDULUM
 
Change of variables

In the link you gave me:
http://api.viglink.com/api/click?fo...M&txt=LARGE-ANGLE MOTION OF A SIMPLE PENDULUM
there is equation 7:
\sin(\varphi)=\frac{\sin(\vartheta/2)}{\sin(\alpha/2)}
It says \alpha changes from 0 to 2\pi for a full oscillation.
First, when \alpha=0 then the denominator=0
Secondly, \alpha changes from -\vartheta to +\vartheta
so \sin(\varphi) changes from -\varphi +\varphi
 
Karol said:
It says \alpha changes from 0 to 2\pi for a full oscillation.
I believe it is \varphi that ranges from 0 to 2\pi for a full oscillation, not \alpha. \alpha is the initial angle of the pendulum, when released from rest.
 
Right, \varphi ranges from 0 to 2\pi
\vartheta is the initial angle
 
Karol said:
Right, \varphi ranges from 0 to 2\pi
\vartheta is the initial angle
No, as Doc Al posted, \alpha is the initial angle. \vartheta varies between -\alpha and +\alpha.
 
Maybe, i don't remember the details now
 

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